Why does the same variable have different values depending on formatting

Question

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    float x=1234.56;
    printf("%10f\n", x);
    return 0;
}

Why does x change value, shouldn't be 1234.560000. It displays 1234.560059

Its because `1234.56` can't be represented exactly in binary. — haccks, Jul 05 '15 at 15:33
Why does the same question get posted over and over again depending on the OP having done zero research whatsoever -1 — Lightness Races in Orbit, Jul 05 '15 at 15:43
possible duplicate of [Is floating point math broken?](http://stackoverflow.com/questions/588004/is-floating-point-math-broken) — ugo, Jul 05 '15 at 16:31

score 0 · Answer 1 · answered Jul 05 '15 at 15:34

0

There's only so many values that can be represented exactly as a float, and 1234.56 ain't one of them:

public class WhatsTheFloatingPoint
{
    public static void main(String[] args)
    {
        float x = 1234.56f;
        BigDecimal y = new BigDecimal(x);
        System.out.println(y);
    }
}

This program prints the closest value that is actually used: 1234.56005859375

answered Jul 05 '15 at 15:34

fredoverflow

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Its C++ tag. Your answer is Java related. BTW, why Java? – haccks Jul 05 '15 at 15:40
@haccks Because Java makes it really easy to see the exact value of a double or a float. I can never remember the appropriate C++ incantations. – fredoverflow Jul 05 '15 at 15:56

score 0 · Answer 2 · answered Jul 05 '15 at 15:36

Its value is always:

1234.560059...

when you show fewer digits, it will round off the tiny extra amount required to map into binary fractions

That is why floating point equivalence is generally done with some acceptable epsilon, or tiny difference.

Note that the amount off can vary by architecture and compiler options

Why does the same variable have different values depending on formatting

2 Answers2